Question: Simplify and expand the following expression: $ \dfrac{3z}{3z - 2}-\dfrac{z - 7}{z + 9} $
Answer: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(3z - 2)(z + 9)$ Multiply the first term by $\dfrac{z + 9}{z + 9}$ $ \begin{align*} \dfrac{3z}{3z - 2} \times \dfrac{z + 9}{z + 9} & = \dfrac{(3z)(z + 9)}{(3z - 2)(z + 9)} \\ & = \dfrac{3z^2 + 27z}{(3z - 2)(z + 9)}\end{align*} $ Multiply the second term by $\dfrac{3z - 2}{3z - 2}$ $ \begin{align*} \dfrac{z - 7}{z + 9} \times \dfrac{3z - 2}{3z - 2} & = \dfrac{(z - 7)(3z - 2)}{(z + 9)(3z - 2)} \\ & = \dfrac{3z^2 - 23z + 14}{(z + 9)(3z - 2)}\end{align*} $ Now we have: $ = \dfrac{3z^2 + 27z}{(3z - 2)(z + 9)} - \dfrac{3z^2 - 23z + 14}{(z + 9)(3z - 2)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{3z^2 + 27z - (3z^2 - 23z + 14)}{(3z - 2)(z + 9)} $ $ = \dfrac{3z^2 + 27z - 3z^2 + 23z - 14}{(3z - 2)(z + 9)} $ $ = \dfrac{50z - 14}{(3z - 2)(z + 9)}$ Expand the denominator: $ = \dfrac{50z - 14}{3z^2 + 25z - 18}$